Is the hypotenuse of a triangle ever divisible by three (for primitive Pythagorean triples)?

Question

Looking for a proof that for primitive Pythagorean triples, the hypotenuse is never divisible by three.

Below are a list of all the primitive Pythagorean triples with a hypotenuses less than 300. None are divisible by 3.

(3, 4, 5) (5, 12, 13) (8, 15, 17) (7, 24, 25) (20, 21, 29) (12, 35, 37) (9, 40, 41) (28, 45, 53) (11, 60, 61) (16, 63, 65) (33, 56, 65) (48, 55, 73) (13, 84, 85) (36, 77, 85) (39, 80, 89) (65, 72, 97) (20, 99, 101) (60, 91, 109) (15, 112, 113) (44, 117, 125) (88, 105, 137) (17, 144, 145) (24, 143, 145) (51, 140, 149) (85, 132, 157) (119, 120, 169) (52, 165, 173) (19, 180, 181) (57, 176, 185) (104, 153, 185) (95, 168, 193) (28, 195, 197) (84, 187, 205) (133, 156, 205) (21, 220, 221) (140, 171, 221) (60, 221, 229) (105, 208, 233) (120, 209, 241) (32, 255, 257) (23, 264, 265) (96, 247, 265) (69, 260, 269) (115, 252, 277) (160, 231, 281) (161, 240, 289) (68, 285, 293)

So the question: is this true for all primitive Pythagorean triples?

Any help would be much appreciated.

Answer 1

The hypotenuse of a primitive triple is never divisible by $3$. For let $(x,y,z)$ be a primitive triple, and suppose $3$ divides $z$. Then $3$ cannot divide $x$ or $y$, else the triple would not be primitive.

It follows that $x^2$ and $y^2$ have remainder $1$ on division by $3$, which means $x^2+y^2$ has remainder $2$ on division by $3$.

Remark: A somewhat more elaborate argument shows that if $p$ is a prime of the form $4k+3$, then $p$ cannot divide the hypotenuse of a primitive triple.

Answer 2

Not significantly different but I'd write this out using the modulus. That is, numbers can be $-1, 0, 1 (\mod 3)$ and thus the squares of them can only be $0, 1$ now $0+0 \equiv 0 (\mod 3)$ so then for a primitive 3-tuple this is not possible; also $1+1 \equiv 2 \equiv - 1(\mod 3)$ and $\nexists c \in \mathbb{N}|c^2 \equiv -1(\mod 3)$. So the only possibility is $a \equiv 0, b\equiv 1, c\equiv 1$ drawing both results in one go: exactly one side is divisible by three in a primitive triple and that side is a leg.

Edit: mod 4 the situation is not different, the squares are still 0 and 1 $(\mod 4)$ and so exactly one side is divisible by four in a primitive triple and that side is a leg.

And it so happens that the smallest set, is an example when these two are different: 3,4,5 and the second 5, 12, 13 is when these two are the same. Isn't mathematics beautiful?